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Write $S={\Phi }_0$, $u={\Phi }_{+}$, $v={\Phi }_{-}$, $M=M_p$, $P=uv-\zeta$, $x=|S{|}^2/M^2$ and $q=|u{|}^2+|v{|}^2$. Assume $g\ne 0$ and, initially, $\zeta \ne 0$. The supergravity F-term potential isThe canonical Kähler potential gives the identity Kähler metric. The three Kähler covariant derivatives of a superpotential areSubstituting, including the negative term rather than treating the potential as just a sum of squares, gives the exact answerHere $(1+x{)}^2-3x=1-x+x^2$ explains the first coefficient. It is strictly positive for real $x$, since $1-x+x^2=(x-1/2{)}^2+3/4$. Thus this model's scalar potential is nonnegative everywhere.

In the stated limit $q/M^2\ll 1$, the exponential can be replaced by $e^x[1+O(q/M^2)]$. The other displayed terms retain the effects of the Kähler covariant derivatives of a superpotential. Small waterfall fields alone do not justify discarding $\zeta /M^2$ in those derivatives, or expanding in $x$. If also $|\zeta |/M^2\ll 1$ and $x\ll 1$, the leading result reduces to the global F-term scalar potentialNo gauge sector or gauge coupling is supplied, so there is no specified D-term contribution to add.

The supersymmetric vacuum manifold of a bilinear hybrid superpotential isIndeed $W=0$ and all three Kähler covariant derivatives of a superpotential vanish there, so all supergravity auxiliary fields vanish and supersymmetry is unbroken. The nonnegative scalar potential makes this a global minimum. Conversely zero potential requires $uv=\zeta$, and then $|S{|}^{2}q=0$; for $\zeta \ne 0$ this forces $S=0$. With just the supplied fields this is a complex vacuum manifold before imposing any gauge quotient: the product is fixed but the ratio of $u$ and $v$ is not. More precisely, it is an F-flat vacuum manifold whether or not a gauge symmetry is specified. If one additionally gauges opposite charges, D-flatness typically imposes $|u|=|v|$, giving $|u|=|v|=\sqrt{|\zeta |}$ up to phases and gauge equivalence; that is an extra assumption, not a consequence of $K$ and $W$ alone.

The other familiar branch of F-term hybrid inflation has $u=v=0$. Its nonzero derivative $D_{S}W=-g\zeta (1+x)$ establishes supersymmetry breaking, and its exact scalar potential isThe canonical-Kähler correction to an F-term hybrid valley begins asIn particular the inflaton quadratic mass cancels at this order; the first lift of the global flat valley is quartic in $|S|$.

To identify when this branch is a transverse minimum, rephase the waterfall fields so that $\zeta$ is real positive. To quadratic order in $u,v$, the global scalar potential isThe normalized combinations $(u\pm \bar{v})/\sqrt{2}$ diagonalize this quadratic form, withThus $|S{|}^2>|\zeta |$ gives a valley of transverse minima, while below the threshold the negative eigenvalue produces the waterfall instability in F-term hybrid inflation.

The exact supergravity correction to this stability test is also available without taking $x$ small. Put $b=|\zeta |/M^2$. Expansion of the exact supergravity F-term potential about $u=v=0$ gives equal diagonal coefficients $|g{|}^2{M}^2{e}^x[x+b^2(1+x^2)]$ and off-diagonal magnitude $|g{|}^2{M}^2{e}^{x}b(1+x+x^2)$. HenceFor the usual sub-Planckian symmetry-breaking scale $0<b\ll 1$, the lower eigenvalue is $|g{|}^2{M}^2{e}^x(1-b)[x-b(1+x^2)]$. It is positive for the familiar sub-Planckian valley $b\ll x\ll 1$, and changes sign near $x=b$.

The printed assertion of two full minima requires qualification. The broken branch is a transverse inflationary valley, not generally a second stationary minimum of the complete supergravity potential. In factIts only stationary point in $S$ is the origin. For $0<|\zeta |<M^2$ that point has a negative waterfall eigenvalue $|g{|}^2(|\zeta {|}^2/M^2-|\zeta |)$. A direct counterexample is the path $S=0$, $u=v=a$ with real small $a$, after rephasing $\zeta >0$:The negative quadratic coefficient proves it is not a minimum. No loop corrections or additional stabilizing interactions were specified that could create the requested second stationary vacuum. The mathematically supported interpretation is a SUSY-breaking transverse valley and a zero-energy supersymmetric vacuum minimum manifold. The limit on $u,v$ alone does not fix the parameter range $|\zeta |/M^2$; the familiar waterfall conclusion uses the sub-Planckian range just stated.

In the corresponding global theory the exponential prefactor, Kähler corrections to $D_{i}W$ and universal $-3|W{|}^2/M^2$ term are absent. Its $u=v=0$ valley has exactly constant tree-level energy $|g\zeta {|}^2$ for all $S$, whereas canonical supergravity lifts it by the quartic correction above. The supersymmetric vacuum $S=0$, $uv=\zeta$ and its zero energy persist in both descriptions. If $\zeta =0$, the origin and the $S$ axis instead have vanishing F-terms; the broken-valley interpretation does not apply.